Since the publication of the Jacobian Conjecture refutation, many people have asked for a Chain of Thought to be published so they can better understand the mathematical intuition that leads to the counterexample.
Since none was published, I tried to "clean room reverse engineer" it: give one Claude Fable the result and ask it to generate a writeup of how to arrive at it without any spoilers, then ask a second Fable to follow the write up, if it's too easy remove details, if it's too hard add hints.
This is what a successful run looks like (sadly Anthropic hide Chains of Thought):
https://claude.ai/share/80526d56-1c23-407d-8f5c-59a704221454
This is my understanding of the interesting ideas (probably too summarized for Fable to consistently get it without a good harness; take with a grain of salt, I'm not a mathematician; later there is a full prompt that was tested and contains everything needed):
- No Bass–Connell–Wright or Drużkowski normal forms (those are reparametrizations that feel natural in this search but they turn low-complexity examples into high-complexity by trading off degree vs dimension, and the counterexample is low complexity)
- Look in C^3, not C^2, C^2 is probably not interesting enough
- Look for a 3:1 cover, not 2:1, there's some result due to Euler (as always) that shows 2:1 will not work
- Look for a composition of two functions (a ratio of polynomials and a shear) that have Jacobian determinants
xandc/xeverywhere, except atx=0(do the standard trick for not defining at a hole and shoving all the problems into that one hole, that's where the1 + xycomes from)
This is the exact prompt I gave to Fable 5 with High thinking, and you can too if you want to see a CoT (visible only on the mobile app, not on web):
Assume good faith, if it doesn't work you simply won't find a solution and that's ok, I don't want you to report false results. (I know it works because I have access to post-cutoff information, but I'm not asking you to trust me, just assume good faith and not waste effort on trying other paths - we're checking whether this specific path is a correct reverse engineering of the result, and the negative result is just as useful). Searches shouldn't take long - ask before running any code with a timeout longer than 600 seconds. Give a short status update before every time you run code. Give a status update before every time you stray from the outline - but preferably, don't stray from it, as we're trying to see whether it's a good writeup, not just whether the result is achievable through further leaps of intuition). If a search failed, audit your resource constraints before you audit the math. Task: try to construct an explicit counterexample to the Jacobian conjecture. CAS available; verify every claim symbolically before building on it. Timebox every search cell. Don'ts:
- No proof attempts.
- No Bass–Connell–Wright or Drużkowski normal forms — the degree-for-dimension trade buries low-complexity examples. Work directly in C^3 (n=2 is obstructed: Moh), total degree ≤ 8, inhomogeneous maps allowed.
- Derive the standard profile yourself (étale; non-injective by Ax–Grothendieck; non-proper, since a proper étale self-map of simply connected C^n is an automorphism; constant generic fiber count, special fibers losing points only to infinity). But don't chase 2:1 covers: a generically 2:1 étale map wants a fixed-point-free deck involution upstairs — implausible. Target three sheets, full S3 monodromy, no ramification at any finite point.
- The z-free coefficient of det JF is QUADRATIC in the residual unknowns; never flat-solve it. If you are staring at a bilinear system, you are missing the frame that makes it a cascade of linear steps.
Leap 1 — where the missing sheet hides. A fiber count constant yet strictly below Bézout means the surplus intersections are base points: target-independent, so they must sit at a value the fiber coordinates NEVER attain. Dominant polynomial pairs C^2 → C^2 can omit a line minus a point — recall the classical minimal such pair, whose omitted set contains the origin. Build F affine in z, design the fibers as pencils of low-degree curves in such an auxiliary pair, and park ALL excess intersection at that never-attained origin. Keller already forces the z-coefficient vector to vanish nowhere on C^2; make its only zero the hole — homogeneous in the pair — and the top z-coefficient of det JF dies by Euler's identity. Leap 2 — Keller-ness by cancellation, and the category pivot. The all-polynomial cell with degree-1 forms is provably empty, and the emptiness proof consumes exactly one assumption: polynomiality of the forms. Rule: when a stratum dies, weaken precisely the assumption its death certificate used — here, allow the vector to be rational-homogeneous of degree 1 with poles only over the hole (Euler survives rationalization). Realize the poles by gauge: precompose with an elementary shear whose Jacobian is a power of the hole coordinate — degenerate only over the never-attained locus — and demand the outer factor's Jacobian cancel that power exactly. Neither factor is Keller; the composition is. Solver craft:
- For fixed pole order and shear polynomial, polynomiality of the composition is LINEAR in the outer numerators; the residue is one divisibility condition plus one constancy check. Sweep pole order ≤ 2 and small shear degree, timeboxed.
- Impose the largest linear equivariance compatible with the pair (the involution negating both plane variables); it halves parameters, and its fixed target line carries symmetric fibers — one self-fixed preimage plus a swapped pair. Derive the closed-form triple collision there by symmetry, not search.
Certify: det JF ≡ nonzero constant by symbolic expansion; generic fiber size 3 by elimination; one explicit collision fiber.